Termination Proof using AProVE

Term Rewriting System R:
[x, y]
a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) -> A(b(a(b(a(x)))))
A(a(f(x, y))) -> A(b(a(x)))
A(a(f(x, y))) -> A(x)
A(a(f(x, y))) -> A(b(a(b(a(y)))))
A(a(f(x, y))) -> A(b(a(y)))
A(a(f(x, y))) -> A(y)
F(a(x), a(y)) -> A(f(x, y))
F(a(x), a(y)) -> F(x, y)
F(b(x), b(y)) -> F(x, y)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

F(b(x), b(y)) -> F(x, y)
F(a(x), a(y)) -> F(x, y)
A(a(f(x, y))) -> A(y)
A(a(f(x, y))) -> A(x)
F(a(x), a(y)) -> A(f(x, y))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

F(a(x), a(y)) -> A(f(x, y))

two new Dependency Pairs are created:

F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(b(x'')), a(b(y''))) -> A(b(f(x'', y'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Forward Instantiation Transformation

Dependency Pairs:

A(a(f(x, y))) -> A(y)
A(a(f(x, y))) -> A(x)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(x), a(y)) -> F(x, y)
F(b(x), b(y)) -> F(x, y)

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

A(a(f(x, y))) -> A(x)

one new Dependency Pair is created:

A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(x), b(y)) -> F(x, y)
F(a(x), a(y)) -> F(x, y)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) -> A(y)

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

A(a(f(x, y))) -> A(y)

two new Dependency Pairs are created:

A(a(f(x, a(f(x'', y''))))) -> A(a(f(x'', y'')))
A(a(f(x, a(f(a(f(x'''', y'''')), y''))))) -> A(a(f(a(f(x'''', y'''')), y'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

A(a(f(x, a(f(a(f(x'''', y'''')), y''))))) -> A(a(f(a(f(x'''', y'''')), y'')))
A(a(f(x, a(f(x'', y''))))) -> A(a(f(x'', y'')))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(x), b(y)) -> F(x, y)
F(a(x), a(y)) -> F(x, y)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(a(x), a(y)) -> F(x, y)

three new Dependency Pairs are created:

F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
F(a(b(x'')), a(b(y''))) -> F(b(x''), b(y''))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(b(x'')), a(b(y''))) -> F(b(x''), b(y''))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, a(f(x'', y''))))) -> A(a(f(x'', y'')))
A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(x), b(y)) -> F(x, y)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, a(f(a(f(x'''', y'''')), y''))))) -> A(a(f(a(f(x'''', y'''')), y'')))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(b(x), b(y)) -> F(x, y)

four new Dependency Pairs are created:

F(b(b(x'')), b(b(y''))) -> F(b(x''), b(y''))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(a(b(x'')), a(b(y''))) -> F(b(x''), b(y''))

four new Dependency Pairs are created:

F(a(b(b(x''''))), a(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(b(b(x'')), b(b(y''))) -> F(b(x''), b(y''))
F(a(b(b(x''''))), a(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, a(f(x'', y''))))) -> A(a(f(x'', y'')))
A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, a(f(a(f(x'''', y'''')), y''))))) -> A(a(f(a(f(x'''', y'''')), y'')))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(b(b(x'')), b(b(y''))) -> F(b(x''), b(y''))

four new Dependency Pairs are created:

F(b(b(b(x''''))), b(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))
F(b(b(a(a(x'''''')))), b(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(b(a(b(x'''''')))), b(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(b(a(a(a(x''''''''))))), b(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

Dependency Pairs:

A(a(f(x, a(f(a(f(x'''', y'''')), y''))))) -> A(a(f(a(f(x'''', y'''')), y'')))
A(a(f(x, a(f(x'', y''))))) -> A(a(f(x'', y'')))
A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))
F(b(b(a(a(a(x''''''''))))), b(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(b(b(a(b(x'''''')))), b(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(b(b(a(a(x'''''')))), b(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(b(b(x''''))), b(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))
F(a(b(b(x''''))), a(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

A(a(f(x, a(f(a(f(x'''', y'''')), y''))))) -> A(a(f(a(f(x'''', y'''')), y'')))
A(a(f(x, a(f(x'', y''))))) -> A(a(f(x'', y'')))
A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(b(x₁)) = x₁

POL(a(x₁)) = x₁

POL(f(x₁, x₂)) = 1 + x₁ + x₂

POL(A(x₁)) = x₁

POL(F(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(b(b(a(a(a(x''''''''))))), b(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(b(b(a(b(x'''''')))), b(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(b(b(a(a(x'''''')))), b(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(b(b(x''''))), b(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))
F(a(b(b(x''''))), a(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(a(b(b(x''''))), a(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))

four new Dependency Pairs are created:

F(a(b(b(b(x'''''')))), a(b(b(b(y''''''))))) -> F(b(b(b(x''''''))), b(b(b(y''''''))))
F(a(b(b(a(a(x''''''''))))), a(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(a(b(b(a(b(x''''''''))))), a(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(a(b(b(a(a(a(x'''''''''')))))), a(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(a(b(b(a(a(a(x'''''''''')))))), a(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(a(b(b(a(b(x''''''''))))), a(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(a(b(b(a(a(x''''''''))))), a(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(b(b(a(b(x'''''')))), b(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(b(a(a(x'''''')))), b(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(b(b(x''''))), b(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))
F(a(b(b(b(x'''''')))), a(b(b(b(y''''''))))) -> F(b(b(b(x''''''))), b(b(b(y''''''))))
F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(b(b(a(a(a(x''''''''))))), b(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(b(b(b(x''''))), b(b(b(y'''')))) -> F(b(b(x'''')), b(b(y'''')))

four new Dependency Pairs are created:

F(b(b(b(b(x'''''')))), b(b(b(b(y''''''))))) -> F(b(b(b(x''''''))), b(b(b(y''''''))))
F(b(b(b(a(a(x''''''''))))), b(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(b(b(b(a(b(x''''''''))))), b(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(b(b(b(a(a(a(x'''''''''')))))), b(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 11

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(a(b(b(a(b(x''''''''))))), a(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(a(b(b(a(a(x''''''''))))), a(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(b(b(b(a(a(a(x'''''''''')))))), b(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(b(b(a(b(x'''''')))), b(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(b(b(a(b(x''''''''))))), b(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(b(b(a(a(a(x''''''''))))), b(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(b(b(b(a(a(x''''''''))))), b(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(b(b(b(b(x'''''')))), b(b(b(b(y''''''))))) -> F(b(b(b(x''''''))), b(b(b(y''''''))))
F(a(b(b(b(x'''''')))), a(b(b(b(y''''''))))) -> F(b(b(b(x''''''))), b(b(b(y''''''))))
F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(b(b(a(a(x'''''')))), b(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(a(b(b(a(a(a(x'''''''''')))))), a(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(a(b(b(b(x'''''')))), a(b(b(b(y''''''))))) -> F(b(b(b(x''''''))), b(b(b(y''''''))))

four new Dependency Pairs are created:

F(a(b(b(b(b(x''''''''))))), a(b(b(b(b(y'''''''')))))) -> F(b(b(b(b(x'''''''')))), b(b(b(b(y'''''''')))))
F(a(b(b(b(a(a(x'''''''''')))))), a(b(b(b(a(a(y''''''''''))))))) -> F(b(b(b(a(a(x''''''''''))))), b(b(b(a(a(y''''''''''))))))
F(a(b(b(b(a(b(x'''''''''')))))), a(b(b(b(a(b(y''''''''''))))))) -> F(b(b(b(a(b(x''''''''''))))), b(b(b(a(b(y''''''''''))))))
F(a(b(b(b(a(a(a(x''''''''''''))))))), a(b(b(b(a(a(a(y'''''''''''')))))))) -> F(b(b(b(a(a(a(x'''''''''''')))))), b(b(b(a(a(a(y'''''''''''')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 12

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(a(b(b(b(a(a(a(x''''''''''''))))))), a(b(b(b(a(a(a(y'''''''''''')))))))) -> F(b(b(b(a(a(a(x'''''''''''')))))), b(b(b(a(a(a(y'''''''''''')))))))
F(a(b(b(b(a(b(x'''''''''')))))), a(b(b(b(a(b(y''''''''''))))))) -> F(b(b(b(a(b(x''''''''''))))), b(b(b(a(b(y''''''''''))))))
F(a(b(b(b(a(a(x'''''''''')))))), a(b(b(b(a(a(y''''''''''))))))) -> F(b(b(b(a(a(x''''''''''))))), b(b(b(a(a(y''''''''''))))))
F(b(b(b(a(a(a(x'''''''''')))))), b(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(b(b(b(a(b(x''''''''))))), b(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(b(b(b(a(a(x''''''''))))), b(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(b(b(b(b(x'''''')))), b(b(b(b(y''''''))))) -> F(b(b(b(x''''''))), b(b(b(y''''''))))
F(a(b(b(b(b(x''''''''))))), a(b(b(b(b(y'''''''')))))) -> F(b(b(b(b(x'''''''')))), b(b(b(b(y'''''''')))))
F(a(b(b(a(a(a(x'''''''''')))))), a(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(b(b(a(a(a(x''''''''))))), b(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(b(b(a(a(x'''''')))), b(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(a(b(b(a(a(x''''''''))))), a(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(b(b(a(b(x'''''')))), b(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(a(b(b(a(b(x''''''''))))), a(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(b(b(b(b(x'''''')))), b(b(b(b(y''''''))))) -> F(b(b(b(x''''''))), b(b(b(y''''''))))

four new Dependency Pairs are created:

F(b(b(b(b(b(x''''''''))))), b(b(b(b(b(y'''''''')))))) -> F(b(b(b(b(x'''''''')))), b(b(b(b(y'''''''')))))
F(b(b(b(b(a(a(x'''''''''')))))), b(b(b(b(a(a(y''''''''''))))))) -> F(b(b(b(a(a(x''''''''''))))), b(b(b(a(a(y''''''''''))))))
F(b(b(b(b(a(b(x'''''''''')))))), b(b(b(b(a(b(y''''''''''))))))) -> F(b(b(b(a(b(x''''''''''))))), b(b(b(a(b(y''''''''''))))))
F(b(b(b(b(a(a(a(x''''''''''''))))))), b(b(b(b(a(a(a(y'''''''''''')))))))) -> F(b(b(b(a(a(a(x'''''''''''')))))), b(b(b(a(a(a(y'''''''''''')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 13

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(a(b(b(b(a(b(x'''''''''')))))), a(b(b(b(a(b(y''''''''''))))))) -> F(b(b(b(a(b(x''''''''''))))), b(b(b(a(b(y''''''''''))))))
F(a(b(b(b(a(a(x'''''''''')))))), a(b(b(b(a(a(y''''''''''))))))) -> F(b(b(b(a(a(x''''''''''))))), b(b(b(a(a(y''''''''''))))))
F(b(b(b(b(a(a(a(x''''''''''''))))))), b(b(b(b(a(a(a(y'''''''''''')))))))) -> F(b(b(b(a(a(a(x'''''''''''')))))), b(b(b(a(a(a(y'''''''''''')))))))
F(b(b(b(a(b(x''''''''))))), b(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(b(b(b(b(a(b(x'''''''''')))))), b(b(b(b(a(b(y''''''''''))))))) -> F(b(b(b(a(b(x''''''''''))))), b(b(b(a(b(y''''''''''))))))
F(b(b(b(a(a(a(x'''''''''')))))), b(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(b(b(b(b(a(a(x'''''''''')))))), b(b(b(b(a(a(y''''''''''))))))) -> F(b(b(b(a(a(x''''''''''))))), b(b(b(a(a(y''''''''''))))))
F(b(b(b(b(b(x''''''''))))), b(b(b(b(b(y'''''''')))))) -> F(b(b(b(b(x'''''''')))), b(b(b(b(y'''''''')))))
F(a(b(b(b(b(x''''''''))))), a(b(b(b(b(y'''''''')))))) -> F(b(b(b(b(x'''''''')))), b(b(b(b(y'''''''')))))
F(a(b(b(a(a(a(x'''''''''')))))), a(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(b(b(a(b(x'''''')))), b(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(a(b(b(a(b(x''''''''))))), a(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(b(b(a(a(a(x''''''''))))), b(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(a(b(b(a(a(x''''''''))))), a(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(b(b(a(a(x'''''')))), b(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(b(b(a(a(x''''''''))))), b(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(a(b(b(b(a(a(a(x''''''''''''))))))), a(b(b(b(a(a(a(y'''''''''''')))))))) -> F(b(b(b(a(a(a(x'''''''''''')))))), b(b(b(a(a(a(y'''''''''''')))))))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(a(b(b(b(b(x''''''''))))), a(b(b(b(b(y'''''''')))))) -> F(b(b(b(b(x'''''''')))), b(b(b(b(y'''''''')))))

four new Dependency Pairs are created:

F(a(b(b(b(b(b(x'''''''''')))))), a(b(b(b(b(b(y''''''''''))))))) -> F(b(b(b(b(b(x''''''''''))))), b(b(b(b(b(y''''''''''))))))
F(a(b(b(b(b(a(a(x''''''''''''))))))), a(b(b(b(b(a(a(y'''''''''''')))))))) -> F(b(b(b(b(a(a(x'''''''''''')))))), b(b(b(b(a(a(y'''''''''''')))))))
F(a(b(b(b(b(a(b(x''''''''''''))))))), a(b(b(b(b(a(b(y'''''''''''')))))))) -> F(b(b(b(b(a(b(x'''''''''''')))))), b(b(b(b(a(b(y'''''''''''')))))))
F(a(b(b(b(b(a(a(a(x'''''''''''''')))))))), a(b(b(b(b(a(a(a(y''''''''''''''))))))))) -> F(b(b(b(b(a(a(a(x''''''''''''''))))))), b(b(b(b(a(a(a(y''''''''''''''))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(a(b(b(b(b(a(a(a(x'''''''''''''')))))))), a(b(b(b(b(a(a(a(y''''''''''''''))))))))) -> F(b(b(b(b(a(a(a(x''''''''''''''))))))), b(b(b(b(a(a(a(y''''''''''''''))))))))
F(a(b(b(b(b(a(b(x''''''''''''))))))), a(b(b(b(b(a(b(y'''''''''''')))))))) -> F(b(b(b(b(a(b(x'''''''''''')))))), b(b(b(b(a(b(y'''''''''''')))))))
F(a(b(b(b(b(a(a(x''''''''''''))))))), a(b(b(b(b(a(a(y'''''''''''')))))))) -> F(b(b(b(b(a(a(x'''''''''''')))))), b(b(b(b(a(a(y'''''''''''')))))))
F(b(b(b(b(a(a(a(x''''''''''''))))))), b(b(b(b(a(a(a(y'''''''''''')))))))) -> F(b(b(b(a(a(a(x'''''''''''')))))), b(b(b(a(a(a(y'''''''''''')))))))
F(b(b(b(b(a(b(x'''''''''')))))), b(b(b(b(a(b(y''''''''''))))))) -> F(b(b(b(a(b(x''''''''''))))), b(b(b(a(b(y''''''''''))))))
F(b(b(b(b(a(a(x'''''''''')))))), b(b(b(b(a(a(y''''''''''))))))) -> F(b(b(b(a(a(x''''''''''))))), b(b(b(a(a(y''''''''''))))))
F(b(b(b(b(b(x''''''''))))), b(b(b(b(b(y'''''''')))))) -> F(b(b(b(b(x'''''''')))), b(b(b(b(y'''''''')))))
F(a(b(b(b(b(b(x'''''''''')))))), a(b(b(b(b(b(y''''''''''))))))) -> F(b(b(b(b(b(x''''''''''))))), b(b(b(b(b(y''''''''''))))))
F(a(b(b(b(a(a(a(x''''''''''''))))))), a(b(b(b(a(a(a(y'''''''''''')))))))) -> F(b(b(b(a(a(a(x'''''''''''')))))), b(b(b(a(a(a(y'''''''''''')))))))
F(b(b(b(a(a(a(x'''''''''')))))), b(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(b(b(b(a(a(x''''''''))))), b(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(a(b(b(b(a(a(x'''''''''')))))), a(b(b(b(a(a(y''''''''''))))))) -> F(b(b(b(a(a(x''''''''''))))), b(b(b(a(a(y''''''''''))))))
F(a(b(b(a(a(a(x'''''''''')))))), a(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(a(b(b(a(b(x''''''''))))), a(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(b(b(a(a(a(x''''''''))))), b(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(b(b(a(a(x'''''')))), b(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(a(b(b(a(a(x''''''''))))), a(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(b(b(a(b(x'''''')))), b(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(b(b(a(b(x''''''''))))), b(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(a(b(b(b(a(b(x'''''''''')))))), a(b(b(b(a(b(y''''''''''))))))) -> F(b(b(b(a(b(x''''''''''))))), b(b(b(a(b(y''''''''''))))))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(b(b(b(b(b(x''''''''))))), b(b(b(b(b(y'''''''')))))) -> F(b(b(b(b(x'''''''')))), b(b(b(b(y'''''''')))))

four new Dependency Pairs are created:

F(b(b(b(b(b(b(x'''''''''')))))), b(b(b(b(b(b(y''''''''''))))))) -> F(b(b(b(b(b(x''''''''''))))), b(b(b(b(b(y''''''''''))))))
F(b(b(b(b(b(a(a(x''''''''''''))))))), b(b(b(b(b(a(a(y'''''''''''')))))))) -> F(b(b(b(b(a(a(x'''''''''''')))))), b(b(b(b(a(a(y'''''''''''')))))))
F(b(b(b(b(b(a(b(x''''''''''''))))))), b(b(b(b(b(a(b(y'''''''''''')))))))) -> F(b(b(b(b(a(b(x'''''''''''')))))), b(b(b(b(a(b(y'''''''''''')))))))
F(b(b(b(b(b(a(a(a(x'''''''''''''')))))))), b(b(b(b(b(a(a(a(y''''''''''''''))))))))) -> F(b(b(b(b(a(a(a(x''''''''''''''))))))), b(b(b(b(a(a(a(y''''''''''''''))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 15

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

F(a(b(b(b(b(a(b(x''''''''''''))))))), a(b(b(b(b(a(b(y'''''''''''')))))))) -> F(b(b(b(b(a(b(x'''''''''''')))))), b(b(b(b(a(b(y'''''''''''')))))))
F(a(b(b(b(b(a(a(x''''''''''''))))))), a(b(b(b(b(a(a(y'''''''''''')))))))) -> F(b(b(b(b(a(a(x'''''''''''')))))), b(b(b(b(a(a(y'''''''''''')))))))
F(b(b(b(b(b(a(a(a(x'''''''''''''')))))))), b(b(b(b(b(a(a(a(y''''''''''''''))))))))) -> F(b(b(b(b(a(a(a(x''''''''''''''))))))), b(b(b(b(a(a(a(y''''''''''''''))))))))
F(b(b(b(b(a(b(x'''''''''')))))), b(b(b(b(a(b(y''''''''''))))))) -> F(b(b(b(a(b(x''''''''''))))), b(b(b(a(b(y''''''''''))))))
F(b(b(b(b(b(a(b(x''''''''''''))))))), b(b(b(b(b(a(b(y'''''''''''')))))))) -> F(b(b(b(b(a(b(x'''''''''''')))))), b(b(b(b(a(b(y'''''''''''')))))))
F(b(b(b(b(a(a(a(x''''''''''''))))))), b(b(b(b(a(a(a(y'''''''''''')))))))) -> F(b(b(b(a(a(a(x'''''''''''')))))), b(b(b(a(a(a(y'''''''''''')))))))
F(b(b(b(b(b(a(a(x''''''''''''))))))), b(b(b(b(b(a(a(y'''''''''''')))))))) -> F(b(b(b(b(a(a(x'''''''''''')))))), b(b(b(b(a(a(y'''''''''''')))))))
F(b(b(b(b(b(b(x'''''''''')))))), b(b(b(b(b(b(y''''''''''))))))) -> F(b(b(b(b(b(x''''''''''))))), b(b(b(b(b(y''''''''''))))))
F(a(b(b(b(b(b(x'''''''''')))))), a(b(b(b(b(b(y''''''''''))))))) -> F(b(b(b(b(b(x''''''''''))))), b(b(b(b(b(y''''''''''))))))
F(a(b(b(b(a(a(a(x''''''''''''))))))), a(b(b(b(a(a(a(y'''''''''''')))))))) -> F(b(b(b(a(a(a(x'''''''''''')))))), b(b(b(a(a(a(y'''''''''''')))))))
F(b(b(b(a(b(x''''''''))))), b(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(a(b(b(b(a(b(x'''''''''')))))), a(b(b(b(a(b(y''''''''''))))))) -> F(b(b(b(a(b(x''''''''''))))), b(b(b(a(b(y''''''''''))))))
F(b(b(b(a(a(a(x'''''''''')))))), b(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(a(b(b(b(a(a(x'''''''''')))))), a(b(b(b(a(a(y''''''''''))))))) -> F(b(b(b(a(a(x''''''''''))))), b(b(b(a(a(y''''''''''))))))
F(a(b(b(a(a(a(x'''''''''')))))), a(b(b(a(a(a(y''''''''''))))))) -> F(b(b(a(a(a(x''''''''''))))), b(b(a(a(a(y''''''''''))))))
F(b(b(a(b(x'''''')))), b(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(a(b(b(a(b(x''''''''))))), a(b(b(a(b(y'''''''')))))) -> F(b(b(a(b(x'''''''')))), b(b(a(b(y'''''''')))))
F(b(b(a(a(a(x''''''''))))), b(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(a(b(b(a(a(x''''''''))))), a(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(a(b(a(a(a(x''''''''))))), a(b(a(a(a(y'''''''')))))) -> F(b(a(a(a(x'''''''')))), b(a(a(a(y'''''''')))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(a(b(a(b(x'''''')))), a(b(a(b(y''''''))))) -> F(b(a(b(x''''''))), b(a(b(y''''''))))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(a(b(a(a(x'''''')))), a(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(b(b(a(a(x'''''')))), b(b(a(a(y''''''))))) -> F(b(a(a(x''''''))), b(a(a(y''''''))))
F(b(b(b(a(a(x''''''''))))), b(b(b(a(a(y'''''''')))))) -> F(b(b(a(a(x'''''''')))), b(b(a(a(y'''''''')))))
F(b(b(b(b(a(a(x'''''''''')))))), b(b(b(b(a(a(y''''''''''))))))) -> F(b(b(b(a(a(x''''''''''))))), b(b(b(a(a(y''''''''''))))))
F(a(b(b(b(b(a(a(a(x'''''''''''''')))))))), a(b(b(b(b(a(a(a(y''''''''''''''))))))))) -> F(b(b(b(b(a(a(a(x''''''''''''''))))))), b(b(b(b(a(a(a(y''''''''''''''))))))))

Rules:

a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:53 minutes

POL(b(x₁))	= x₁
POL(a(x₁))	= x₁
POL(f(x₁, x₂))	= 1 + x₁ + x₂
POL(A(x₁))	= x₁
POL(F(x₁, x₂))	= 1 + x₁ + x₂